Question

Is (y+1)^2 - 3 = 2x a function? If it is, what is its domain and range?

Answer 1

ya it is a function
it fefines implicitly f(x) = y in some domain (have to ckeck carefully)
becouse F(x,y) has not vanishing and continuous partial derivative of y.

Answer 2

what?

Answer 3

what u study in, university, highschool?

Answer 4

high school

Answer 5

ok

Answer 6

then u can wtite it like this:
\[((y+1)^{2}-3)/2=x\]
then it becomes function of y, it means x =f(y)
it is well defind function, (inverse of y) with domain all R, and range (-3/2,infinity)

Answer 7

for finding f(x) = y, make change of variable y' = y+1 so your exprecion becomes\[y'^{2} = 2x+3\] which is equal to\[y' =\sqrt{2x +3}\] domain of this is x>-3/2 and range all positive real numbers

Answer 8

ah, forgot one thing: remmeber to come back to your variable y = y' -1, so range will be -1,infinity.

Answer 9

hope it helps

Answer 10

so it goes:

\[(y+1)^{2} - 3 = 2x\]
\[(y+1)^{2}-3-2x = 0\]
\[(y+1)^{2} = 3+2x\]
\[y = -1 +/- \sqrt{3+2x}\]

Answer 11

but what happens again for x?

Answer 12

the exprecion under the root sign can't be negative, so x cant be smaller than -3/2

Answer 13

it means the domain is (-3/2,infinity)